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Who will do the dishes?



To decide who will do the dishes, we put in a jar tickets with prime numbers up to 111. The one who draws the ticket with a larger number wins.

I look at my ticket: 3. Maybe you have a low number like 11... So I offer to exchange our tickets before showing them.

What is the probability I will do the dishes?

Edited by harey
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OK, so here the solution.


If you have the highest number, you will not exchange it. Recursively, we can restart the game without the highest number. You will not agree to exchange as long as you do not have the lowest number.

As I have the second lowest number, my only hope is that you have 2. However, I spoiled my last chance because in this case, you trade.

So I will 100% do the dishes.

Should you have problems with the recursivity:
if you have the 2nd highest number, you will not exchange because you will not receive the highest number
if you have the 3rd highest number, we have seen that higher numbers will not be exchanged


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